Chevalley's centralizer theorem and reductive centralizers

Let $I_u=({\bigcap }_{B\supseteq T}{B}_u{)}_{\mathrm{r}\mathrm{e}\mathrm{d}}^{\circ }$ and $I=({\bigcap }_{B\supseteq T}B{)}_{\mathrm{r}\mathrm{e}\mathrm{d}}^{\circ }$. The intersection set is finite by [F1]. These groups are smooth connected and normalized by $N_G(T)$, which permutes the factors; $I_u$ is unipotent as a subgroup of one $B_u$. The normal unipotent radical $R_u(G)$ lies in every $B$ by [F2] and then in every $B_u$ by its maximal normal-unipotent property, so $R_u(G)\subseteq I_u$. We prove the reverse inclusion by the action on $X=G\mathrm{/}{B}_0$.

We give the closed-orbit argument for unipotent actions on affine varieties. For an orbit $O$, let $Y$ be its reduced affine closure. If its boundary is nonempty, the ideal of the boundary in $O(Y)$ is a nonzero stable rational representation: the orbit is open dense in $Y$, so its boundary is proper. By [F5] choose a nonzero invariant function $f$ in this ideal. Its value is constant on the dense orbit, hence $f$ is that scalar on reduced $Y$. The boundary forces this scalar to be zero, contradicting $f\mathrm{\ne }0$. Therefore every such orbit is closed. This is the Kostant–Rosenlicht argument used in Milne17.65.

Under $G\mathrm{/}{B}_0\to \{\text{Borels}\}$, $g{B}_0$ corresponds to $g{B}_0{g}^{-1}$; the map is injective because $N_G(B_0)=B_0$. Thus $X^T(k)$ is the finite set of Borels containing $T$, and $N_G(T)(k)$ is transitive on it. Take the nondegenerate projective embedding $X\subseteq \mathbf{P}(V)$ from [F4], with weights $\mathrm{\Xi }$ for $T$. Choose an integral cocharacter $\lambda$ pairing distinctly with the distinct elements of $\mathrm{\Xi }$. Let ${\chi }_{-}$ have minimum pairing. On the nonempty open set where the projection to $V_{{\chi }_{-}}$ is nonzero, the limit under $\lambda (t)$ as $t\to 0$ is that projected line, lying in $X\cap \mathbf{P}(V_{{\chi }_{-}})\subseteq X^T$. This projection has constant image because its irreducible domain maps to the finite fixed set. Since $X$ spans $V$, its projections span $V_{{\chi }_{-}}$, so this space is one-dimensional. Write it as $k{v}_{-}$, with point $x_{-}=[v_{-}]\in X^T$.

Let $\mathrm{\ell }\in V^{\ast }$ equal $1$ on $v_{-}$ and vanish on every other weight space. The chart $U(x_{-})=X\cap \{\mathrm{\ell }\mathrm{\ne }0\}$ is affine and is contracted by $\lambda$ to $x_{-}$. In the dual projective space, every $G$-orbit meets the affine chart where evaluation on $v_{-}$ is nonzero: otherwise a nonzero dual vector would annihilate all $g{v}_{-}$, which span $V$. The action of ${\lambda }^{-1}$ contracts that dual chart to $[\mathrm{\ell }]$, so the closure of every dual orbit contains $[\mathrm{\ell }]$. A closed orbit in the closure of $G[\mathrm{\ell }]$ exists by [F4]; it also contains $[\mathrm{\ell }]$, and hence is $G[\mathrm{\ell }]$. Thus $G[\mathrm{\ell }]$ is closed and its stabilizer $P$ has complete quotient. By Borel fixed points [F4], $P$ contains a Borel. Since $T\subseteq P$, choose a Borel containing $T$ inside $(P_{\mathrm{r}\mathrm{e}\mathrm{d}}{)}^{\circ }$; conjugacy there to the previously obtained $G$-Borel shows it is a $G$-Borel $B^{\mathrm{\prime }}$. Thus $I_u\subseteq B_u^{\mathrm{\prime }}\subseteq P$. The dual line is $I_u$-stable, so $U(x_{-})$ is $I_u$-stable.

Translating this chart by $N_G(T)(k)$ supplies a $T$-stable and $I_u$-stable affine open $U(x)$ containing every $x\in X^T(k)$, since this normalizer preserves $I_u$. These charts cover $X$: the closure of the $T$-orbit of any point is nonempty complete and has a $T$-fixed point $x$ by [F4]; if the point lay outside $U(x)$, its full orbit closure would lie in the closed $T$-stable complement, contradicting the presence of $x$.

For any $y\in X(k)$, the complete orbit closure $\overline{I_{u}y}$ has an $I_u$-fixed point $z$ by [F4]. Choose an $I_u$-stable affine chart from step 4.1 containing $z$. If the orbit met its closed stable complement, the whole orbit and its closure would lie there, excluding $z$. Thus the orbit lies in the chart and is closed there by step 1.2. It contains $z$, so it is a single point. Hence $I_u$ fixes every point of $X$. The smooth reduced scheme $I_u\times X$ has dense rational points, so the action is scheme-theoretically trivial. Its stabilizers are all Borels, and consequently $I_u$ lies in their full intersection. Smoothness and connectedness put it in its reduced neutral component $R(G)$ by [F2]; being unipotent it lies in $R_u(G)$. Together with step 1.1 this proves $I_u=R_u(G)$.

Fix $B_0\supseteq T$ and its split quotient $q_0:B_0=B_{0,u}⋊T\to T$. The group $I$ contains $T$ and maps onto $T$ with this section. Thus $I=K⋊T$, with $K=I\cap B_{0,u}$; the product isomorphism shows $K$ smooth connected. It is unipotent and lies in every Borel $B\supseteq T$, so its homomorphism into the torus $B\mathrm{/}{B}_u$ is trivial by [F6]. Therefore $K\subseteq B_u$ for all these Borels, hence $K\subseteq I_u$. Conversely $I_u\subseteq K$, so $K=I_u=R_u(G)$ and $I=R_u(G)T$. This proves both Chevalley intersection identities scheme-theoretically.

For a torus subgroup $S$, set $C=C_G(S)$ and choose a maximal torus $T\supseteq S$. The group $R_u(G)\cap C=R_u(G{)}^S$ is smooth connected by [F3], unipotent and normal in $C$, so it lies in $R_u(C)$. Conversely for every Borel $B\supseteq T$, $C\cap B$ is a Borel of $C$ by [F3], and the normal smooth connected unipotent subgroup $R_u(C)$ lies in it by [F2]. Thus $R_u(C)\subseteq I=R_u(G)T$. Its map into the torus $I\mathrm{/}{R}_u(G)$ is trivial, so $R_u(C)\subseteq R_u(G)\cap C$. Equality follows. For an external torus action form $H=G⋊S$. Its unipotent radical is $R_u(G)$: normal unipotent subgroups have trivial image in $S$, and $R_u(G)$ is invariant under $S$ by uniqueness. Since $C_H(S)=G^S\times S$, the subgroup-centralizer equality in $H$ gives $R_u(G^S)=R_u(G{)}^S$.

If $G$ is reductive, $R_u(G)=1$, so step 7.1 and smooth connectedness in [F3] make every torus fixed subgroup, in particular every torus centralizer, reductive. For maximal $T$, the smooth connected $C_G(T)$ lies in every Borel containing $T$ by [F1], hence in $I=R_u(G)T=T$ by step 6.1; the reverse inclusion is immediate. Therefore $C_G(T)=T$. This proves every stated consequence.